Question

Using the data provided in the excel file, show all of your work for the following calculations:
a.) mean temperature of unmixed reagents (^oC)

b.) δελταT from graph (^oC)

c.) q absorbed by reaction mixture (J)

d.) q absorbed by calorimeter, stirrer, and thermometer (J)

e.) q total absorbed (J)

f.) q total released (J)

g.) calculation to show limiting reagent

h.) deltaH neutralization for the reaction (kJ/mole of acid)

A student reacted 100.0 mL of 0.9800 M HCl with 100.0 mL of 0.9900 M NH3. The density of the reaction mixture was 1.02 g/mL and the heat capacity was 4.016 J/g K. Calculate the enthalpy of neutralization by plotting and using the data shown below.
Data for mixing hydrochloric acid and ammonia
Time(min)	Temp(oC)
0.0	23.25
0.5	23.27
1.0	23.28
1.5	23.30
2.0	23.30
3.0	23.35
4.0	23.44
4.5	23.47
mix	---------
5.5	28.75
6.0	28.50
7.0	28.55
8.0	28.48
9.0	28.32
10.0	28.25
11.0	28.20
12.0	28.05
13.0	27.96
14.0	27.80
15.0	27.75

Answer 1

(a) mean temperature of unmixed reactant = 23.33 ^oC

(b) question is not clear

(c) heat absorbed by the reaction mixture = mass * heat capacity

mass of the reaction mixture = density * volume

                                          = 1.02 g/ mL * 200 mL = 204 g

Mean temperature after mixing = 28.24 ^oC

change in temperature = 28.24- 23.33 = 4.91^oC = 277.91 K

heat absorbed by reaction mixture = 204g * 4.16 J/g K * 277.91= 235.8k J

(d) heat absorbed by the calorimeter = mass of calorimeter * heat capacity * temp change

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Calculation of the limiting reagent:

NH₃ + HCl -------> NH₄Cl

1 mole NH3 reacts with 1 mole of HCl

moles of NH3 present = 0.99*100/1000 = 0.099 mole

moles of HCl present = 0.98 *100/1000 = 0.098 moles

As HCl is present in less amount, It will be consumed completely and it is the limiting reagent.